Square root of a product of negative numbers

Question

I'm curious about a specific rule about the square root of a product of two numbers. Let $a,b>0$. Then $$ \sqrt{ab}=\sqrt{a}\sqrt{b}. $$ We also have that $$ \sqrt{-ab}=\sqrt{-a}\sqrt{b}. $$ However, it is not true that $$ \sqrt{(-a)(-b)}=\sqrt{-a}\sqrt{-b}. $$ Indeed, if it was true, we would get $1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i^2=-1$, a contradiction.

My question is the following: Is there a stronger reason that makes this impossible? I mean, besides leading to a contradiction given what we know from real and complex numbers. Is there a generalization of the definition of root that includes this kind of manipulation? I'm looking for some deeper understanding, any insight is appreciated.

Answer 1

The deeper understanding that prevents a nice generalization of the rule $$ \sqrt{ab} = \sqrt{a}\sqrt{b} $$ that works for nonnegative $a$ and $b$ is the fact that for every nonzero complex number $z$ there are two solutions to the equation $$ x^2 = z . $$ There is no consistent way to pick one of those roots and call it $\sqrt{z}$ unless $z$ happens to be real and positive. In that case that expression always means the positive root. Only on that domain is $\sqrt{\ }$ a function.

It's often convenient to write $\sqrt{-r} = i\sqrt{r} $ for positive real $r$, but that is not a formal definition. Using it as one leads to the contradiction in the question.

Answer 2

The square root operation is more honestly seen as a function which has two possible values. This is best explained by example: $$\begin{aligned} \sqrt{0} &= \{0\} \\ \sqrt{1} &= \{1, -1\} \\ \sqrt{-4} &= \{2i, -2i\} \\ \sqrt{2i} &= \{1 + i, -1 -i \} \end{aligned}$$ If we interpret the square root like this, where "multiplying sets" means taking all possible products, then all the problems go away: $$ \sqrt{1} = \{1, -1\} = \{i, -i\} \cdot \{i, -i\} = \sqrt{-1} \sqrt{-1}$$ but unfortunately I cannot proceed by saying $\sqrt{-1} \sqrt{-1} = i \cdot i = -1$, since $\sqrt{-1}$ is not equal to $i$, it is equal to the set $\{i, -i\}$. If you force a choice of a single value out of the square root, it will break its algebraic properties.

You may pick one and only one from this list:

1. The square root is a function resulting in a single value. Care must be taken when doing algebra under the radical sign.
2. The square root is a function resulting in perhaps many values. You may do algebra under the radical sign freely.

Answer 3

In addition to the other answers, I think part of the "deeper" reason why the "law" $\sqrt{a}\sqrt{b}=\sqrt{ab}$ is that the field of complex numbers is not an ordered field.

Think about it this way: as long as we stick to positive numbers $a$, the two solutions to the equation $x^2=a$ always live in $\mathbb R$, the set of real numbers. $\mathbb R$ is an ordered field: given any two members of $\mathbb R$, you can always say which one comes before and which one comes after the other. This makes it possible to distinguish half of the number line as the "positive ray" and the other half as the "negative ray", which in turn makes it possible to define the symbol $\sqrt{a}$ unambiguously to mean the positive solution to $x^2 = a$, and it is this convention that makes the formula $\sqrt{a}\sqrt{b} = \sqrt{ab}$.

However, the field of complex numbers, $\mathbb C$, is not an ordered field, and in fact there is no meaningful way to make it into one. Given two complex numbers there is no way to say which one is "first" and which one "second"; if you imagine yourself sitting at a point in $\mathbb C$, there is no clear way to say which way is forward and which way is backward. This lack of orientation means that when $a$ is negative, it is impossible to choose which of the two solutions to $x^2 = a$ should be denoted by $\sqrt{a}$, and without being able to unequivocally know which complex number is meant, the expression itself becomes meaningless (or at best ambiguous).

So the reason that $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is not reliable when $a$ is negative is that the symbols themselves, the expressions $\sqrt{a}$ and $\sqrt{b}$, cannot be unambiguously defined, which in turn is because they exist in a field that cannot be given an order relation.

Answer 4

Maybe you should learn some basic complex analysis.

In complex analysis, powers are multivalued. They are defined by $$ z^a=e^{a \operatorname{Log} z} $$ And because $\operatorname{Log}(z)$ is a multi-valued function, the power function is multi-valued too. That explains why there's two answers to your square root.

See This Page for further study.